Question: Find $\lim_{h\to 0}\dfrac{\cos\left(\dfrac{\pi}{6}+h\right)-\cos\left(\dfrac{\pi}{6}\right)}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac12$ (Choice B) B $\dfrac12$ (Choice C) C $2$ (Choice D) D The limit doesn't exist
The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $\cos\left(\dfrac{\pi}{6}+h\right)-\cos\left(\dfrac{\pi}{6}\right)$, we can tell that the function is $f(x)=\cos(x)$ and the $x$ -value is $\dfrac{\pi}{6}$. In other words, the limit expression is equal to $f'\left(\dfrac{\pi}{6}\right)$ for $f(x)=\cos(x)$. Let's find $f'(x)$ : $f'(x)=-\sin(x)$ Now let's evaluate $f'\left(\dfrac{\pi}{6}\right)$ : $f'\left(\dfrac{\pi}{6}\right)=-\sin\left(\dfrac{\pi}{6}\right)=-\dfrac12$ In conclusion, $\lim_{h\to 0}\dfrac{\cos\left(\dfrac{\pi}{6}+h\right)-\cos\left(\dfrac{\pi}{6}\right)}{h}=-\dfrac12$.